Problem Set 1: Hodrick-Prescott Filter

Introduction

A commonly used approach to decompose a time series into a permanent component \(y^p\) and a transitory (cyclical) component \(y^c\) goes back to Hodrick and Prescott (1997). Suppose you have a sequence of data, \(y_t\) with \(t = 1,...,T\) (i.e. there are \(T\) total observations). Our objective is to find a trend \(\{y^p_t\}^T_{t=1}\) to minimize the following objective function:

\[f = \underset{y^p_t}{\min} \sum^T_{t=1} (y_t - y^p_t)^2 + \lambda \sum^{T-1}_{t=2} [(y^p_{t+1} - y^p_t) - (y^p_t - y^p_{t-1})]^2\]

where the parameter \(\lambda \geq 0\).

Exercise 1

Question: Provide a verbal interpretation of this objective function. What are the trade-offs?*

Answer: The first term of the function is the sum of quadratic differences between the realization \(y_t\) and the trend component \(y^p_t\). If the function would only consist of this component, the minimization problem would yield \(y^p_t = y_t\). Therefore, \(y^p_t\) has also to minimize the squared difference to its neighbours \(y^p_{t+1}\) and \(y^p_{t-1}\) so that all trend components are similar to each other. The weighting parameter \(\lambda\) controls how important it is that the trend values are similar to each other. A higher \(\lambda\) will lead to more equal trend components meaning a smoother trend, but it might also wash out important differences in the trend.

Exercise 2

Question: Prove that if \(\lambda = 0\), the solution is \(y^p_t = y_t \forall t\), i.e. the trend and the actual series are identical.

Answer: If \(\lambda = 0\) the former minimization problem simplifies to

\[\underset{y^p_t}{\min} \sum^T_{t=1} (y_t - y^p_t)^2\]

The first derivative with respect to \(y^p_t\) is given by

\[\begin{split}\begin{align} \frac{d f}{d y^p_t} &= 2 (y_t - y^p_t)\\ &= y_t - y^p_t\\ &= 0 \end{align}\end{split}\]

Exercise 3

Question: Prove that as \(\lambda \to \infty\), the filtered series \(\{y^p_t\}^T_{t=1}\) is linear (i.e. \(y^p_t = \beta t\) for some \(\beta\))

Answer: If the formula is divided by \(\lambda\) and \(\lambda \to \infty\) the first term of the equation approaches 0.

Exercise 4

Question: For the more general case in which \(0 < \lambda < \infty\), derive analytical conditions which implicitly define the trend. To do this, take the derivative with respect to \(y^p_t\) for each \(t = 1, \dots, T\) and set them equal to zero, yielding \(T\) first order conditions.

Answer: Here are the three different derivations:

For \(y^p_1\):

\[\begin{split}\begin{align} - (y_1 - y^p_1) + \lambda (y^p_1 - 2 y^p_2 + y^p_3) &= 0\\ (1 + \lambda) y^p_1 - 2 \lambda y^p_2 + \lambda y^p_3 &= y_1 \end{align}\end{split}\]

For \(y^p_2\):

\[\begin{split}\begin{align} - (y_2 - y^p_2) + \lambda [-2(y^p_1 - 2 y^p_2 + y^p_3) + (y^p_2 - 2 y^p_3 + y^p_4)] &= 0\\ -2 \lambda y^p_1 + (1 - 5\lambda) y^p_2 - 4 \lambda y^p_3 + \lambda y^p_4 &= y_2 \end{align}\end{split}\]

For \(y^p_t\) for \(3 \leq t \leq T-2\):

\[\begin{split}\begin{align} - (y_t - y^p_t) + \lambda [(y^p_t - 2y^p_{t-1} + y^p_{t-2}) - 2(y^p_{t+1} - 2 y^p_t + y^p_{t-1}) + (y^p_{t+2} - 2y^p_{t+1} + y^p_{t}) &= 0\\ \lambda y^p_{t-2} - 4 \lambda y^p_{t-1} + (1 + 6 \lambda) y^p_t - 4 \lambda y^p_{t+1} + \lambda y^p_{t+2} &= y_t \end{align}\end{split}\]

for \(y^p_{T-1}\):

\[\begin{split}\begin{align} - (y_{T-1} - y^p_{T-1}) + \lambda [2(y^p_{T-3} - 2y^p_{T-2} + y^p_{T-1}) - 2(y^p_T - 2y^p_{T-1} + y^p_{T-2})] &= 0\\ \lambda y^p_{T-3} - 4 \lambda y^p_{T-2} + (1 + 5\lambda) y^p_{T-1} - 2\lambda y^p_T&= y_{T-1} \end{align}\end{split}\]

for \(y^p_{T}\):

\[\begin{split}\begin{align} - (y_T - y^p_T) + \lambda(y^p_T - 2y^p_{T-1} + y^p_{T-2}) &= 0\\ \lambda y^p_{T-2} - 2\lambda y^p_{T-1} + 2\lambda y^p_T &= y_T \end{align}\end{split}\]

Exercise 5

Question: Write a Julia function to find the trend taking the actual data series \(y_t\) and parameter \(\lambda\) as inputs. To do this, express the first order conditions in Exercise 4 in matrix form as follows:

\[\begin{split}\begin{align} \Lambda Y^t &= Y\\ Y^t &= \Lambda^{-1}Y \end{align}\end{split}\]

where \(\Lambda\) is a \(T \times T\) matrix whose elements are functions of \(\lambda\), \(Y^t\) is a \(T \times 1\) vector equal to \([y^t_1, y^t_2, \dots, y^t_T]'\) and \(Y\) is a \(T \times 1\) vector equal to \([y_1, y_2, \dots, y_T]'\).

function hodrick_prescott_filter(y, lambda)
    T = size(y, 1)
    matrix = zeros(T, T)

    matrix[1, 1:3] = [1 + lambda, -2 * lambda, lambda]
    matrix[2, 1:4] = [-2 * lambda, 1 + 5 * lambda, -4 * lambda, lambda]

    for i = 3 : T - 2
        matrix[i, i-2 : i+2] = [lambda, -4*lambda, 1 + 6 * lambda, -4 * lambda, lambda]
    end

    matrix[T-1, T-3:T] = [lambda, -4 * lambda, 1 + 5 * lambda, -2 * lambda]
    matrix[T, T-2:T] = [lambda, -2 * lambda, 1 + lambda]

    trend = matrix \ y
    cycle = y - trend

    return trend, cycle
end;

Exercise 6

using DataFrames, Plots, SparseArrays, XLSX

y = float.(XLSX.readdata("problem_set_1_data/us_real_gdp.xlsx", "FRED Graph", "B12:B295"));

trend, cycle = hodrick_prescott_filter(y, 1600);

timeline = 1947:0.25:2017.75;

plot(timeline, cycle)

plot(timeline, trend)

To validate the method, we take another implementation of the HP-Filter from http://www.econforge.org/posts/2014/juil./28/cef2014-julia/ (note the API change introduced by https://github.com/JuliaLang/julia/pull/23757 and the new form explained in https://github.com/JuliaLang/julia/pull/23757/files#diff-7904f4ddd9158030529e0ed5ee8707eeR1771).

function hp_filter(y, lambda)
    n = length(y)
    @assert n >= 4

    diag2 = lambda*ones(n-2)
    diag1 = [ -2lambda; -4lambda*ones(n-3); -2lambda ]
    diag0 = [ 1+lambda; 1+5lambda; (1+6lambda)*ones(n-4); 1+5lambda; 1+lambda ]

    D = spdiagm(-2 => diag2, -1 => diag1, 0 => diag0, 1 => diag1, 2 => diag2)

    trend = D \ y
    cycle = y - trend

    return trend, cycle
end;

trend, cycle = hp_filter(y, 1600);

plot(timeline, cycle)

plot(timeline, trend)